Differentiation of Inverse Trigonometric Function (cos/sine/tan) - Practice Questions & MCQ

Differentiation of Inverse Trigonometric Function (sin/cos/tan)

14. $\frac{d}{dx}\left(\sin^{-1}(\mathrm{x})\right) = \frac{1}{\sqrt{1-\mathrm{x}^2}}$

Let $\sin^{-1} \mathrm{x} = \mathrm{y}$ where $\mathrm{y} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

\[
\begin{aligned}
\therefore & \quad x = \sin y, \\
& \Rightarrow \frac{dx}{dy} = \cos y = \sqrt{1-\mathrm{x}^2} \quad (\because \cos y \geq 0 \quad \forall y \in [-\pi/2, \pi/2]), \\
& \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1-\mathrm{x}^2}}.
\end{aligned}
\]

15. $\frac{d}{dx}\left(\cos^{-1}(\mathrm{x})\right) = -\frac{1}{\sqrt{1-\mathrm{x}^2}}$

16. $\frac{d}{dx}\left(\tan^{-1}(\mathrm{x})\right) = \frac{1}{1+\mathrm{x}^2}$

Let $\tan^{-1} \mathrm{x} = \mathrm{y}$ where $\mathrm{y} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

\[
\begin{aligned}
\therefore & \quad x = \tan y, \\
& \Rightarrow \frac{dx}{dy} = \sec^2 y, \\
& \Rightarrow \sec^2 y = 1+\tan^2 y, \\
& \Rightarrow 1+\tan^2 y = 1+\mathrm{x}^2, \\
& \Rightarrow \frac{dy}{dx} = \frac{1}{1+\mathrm{x}^2}.
\end{aligned}
\]

Differentiation of Inverse Trigonometric Function (CSC/sec/cot)

17. $\quad \frac{d}{d x}\left(\cot ^{-1}(\mathrm{x})\right)=-\frac{1}{1+\mathbf{x}^2}$

As $\cot ^{-1}(x)=\frac{\pi}{2}-\tan ^{-1}(x)$
Differentiating both sides we get,

$
\frac{d\left(\cot ^{-1}(x)\right)}{d x}=-\frac{1}{1+x^2}
$

18. $\quad \frac{d}{d x}\left(\sec ^{-1}(\mathbf{x})\right)=\frac{1}{|\mathbf{x}| \sqrt{\mathbf{x}^2-1}}$
19. $\frac{d}{d x}\left(\csc ^{-1}(\mathbf{x})\right)=-\frac{1}{|\mathbf{x}| \sqrt{\mathbf{x}^2-1}}$

Substitutions for Inverse trigonometric functions

\[
\begin{array}{|c|c|c|}
\hline 
\text{S. No.} & \text{Expression} & \text{Substitutions} \\ 
\hline 
1. & \sqrt{a^2-x^2} & x = a \sin \theta \text{ or } a \cos \theta \\ 
\hline 
2. & \sqrt{a^2+x^2} & x = a \tan \theta \text{ or } a \cot \theta \\ 
\hline 
3. & \sqrt{x^2-a^2} & x = a \sec \theta \text{ or } a \csc \theta \\ 
\hline 
4. & \sqrt{\frac{a+x}{a-x}} \text{ or } \sqrt{\frac{a-x}{a+x}} & x = a \cos \theta \text{ or } a \cos 2\theta \\ 
\hline 
5. & \sqrt{a+x} \text{ or } \sqrt{a-x} & x = a \cos \theta \text{ or } a \cos 2\theta \\ 
\hline 
\end{array}
\]